If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma },\mathrm{\sigma }$ are the roots of the equation ${\mathrm{x}}^4+4{\mathrm{x}}^3-6{\mathrm{x}}^2+7\mathrm{x}-9=0$,then the value of $\left(1+{\mathrm{\alpha }}^2\right)\left(1+{\mathrm{\beta }}^2\right)\left(1+{\mathrm{\gamma }}^2\right)\left(1+{\mathrm{\sigma }}^2\right)$ is

Since $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma },\mathrm{\sigma }$ are the roots of the given equation, we have ${\mathrm{x}}^4+4{\mathrm{x}}^3-6{\mathrm{x}}^2+7\mathrm{x}-9=(\mathrm{x}-\mathrm{\alpha })(\mathrm{x}-\mathrm{\beta })(\mathrm{x}-\mathrm{\gamma })(\mathrm{x}-\mathrm{\sigma })$
Putting x = i and then x = -i, we get
$1-4\mathrm{i}+6+7\mathrm{i}-9=(\mathrm{i}-\mathrm{\alpha })(\mathrm{i}-\mathrm{\beta })(\mathrm{i}-\mathrm{\gamma })(\mathrm{i}-\mathrm{\sigma })$ and $1+4\mathrm{i}+6-7\mathrm{i}-9=(-\mathrm{i}-\mathrm{\alpha })(-\mathrm{i}-\mathrm{\beta })(-\mathrm{i}-\mathrm{\gamma })(-\mathrm{i}-\mathrm{\sigma })$
Multiplying these two equations, we get 
$(-2+3\mathrm{i})(-2-3\mathrm{i})=\left(1+{\mathrm{\alpha }}^2\right)\left(1+{\mathrm{\beta }}^2\right)\left(1+{\mathrm{\gamma }}^2\right)\left(1+{\mathrm{\sigma }}^2\right)$ or   $13=\left(1+{\mathrm{\alpha }}^2\right)\left(1+{\mathrm{\beta }}^2\right)\left(1+{\mathrm{\gamma }}^2\right)\left(1+{\mathrm{\sigma }}^2\right)$